Fall 07 PHYS 3 Chapter 9: 5, 0,, 3, 3, 34 5. ssm The drawing shows a jet engine suspended beneath the wing of an airplane. The weight W of the engine is 0 00 N and acts as shown in the drawing. In flight the engine produces a thrust T of 6 300 N that is parallel to the ground. The rotational axis in the drawing is perpendicular to the plane of the paper. With respect to this axis, find the magnitude of the torque due to (a) the weight and (b) the thrust. Problem 5 REASONING To calculate the torques, we need to determine the lever arms for each of the forces. These lever arms are shown in the following drawings:.50 m 3.0 W W = (.50 m) sin 3.0 T SOLUTION a. Using Equation 9., we find that the magnitude of the torque due to the weight W is W Magnitude of torque = W = 0 00 N.5 m sin 3 = 3 500 N m b. Using Equation 9., we find that the magnitude of the torque due to the thrust T is T.50 m 3.0 T = (.50 m) cos 3.0 Magnitude of torque = T = 6 300 N.5 m cos 3 = 3 000 N m
Fall 07 PHYS 3 Chapter 9: 5, 0,, 3, 3, 34 **0. A rotational axis is directed perpendicular to the plane of a square and is located as shown in the drawing. Two forces, F and F, are applied to diagonally opposite corners, and act along the sides of the square, first as shown in part a and then as shown in part b of the drawing. In each case the net torque produced by the forces is zero. The square is one meter on a side, and the magnitude of F is three times that of F. Find the distances a and b that locate the axis. REASONING AND SOLUTION The net torque about the axis in text drawing (a) is Στ = τ + τ = F b F a = 0 Considering that F = 3F, we have b 3a = 0. The net torque in drawing (b) is then Στ = F (.00 m a) F b = 0 or.00 m a 3b = 0 Solving the first equation for b, substituting into the second equation and rearranging, gives a = 0.00 m and b = 0.300 m
Fall 07 PHYS 3 Chapter 9: 5, 0,, 3, 3, 34. The drawing shows a person (weight, W = 584 N) doing push-ups. Find the normal force exerted by the floor on each hand and each foot, assuming that the person holds this position. REASONING The drawing shows the forces acting on the person. It also shows the lever arms for a rotational axis perpendicular to the plane of the paper at the place where the person s toes touch the floor. Since the person is in equilibrium, the sum of the forces must be zero. Likewise, we know that the sum of the torques must be zero. SOLUTION Taking upward to be the positive direction, we have F + F W = 0 FEET HANDS Remembering that counterclockwise torques are positive and using the axis and the lever arms shown in the drawing, we find Substituting this value into the balance-of-forces equation, we find F = W F = 584 N 39 N = 9 N FEET HANDS The force on each hand is half the value calculated above, or 96 N. Likewise, the force on each foot is half the value calculated above, or 96 N.
Fall 07 PHYS 3 Chapter 9: 5, 0,, 3, 3, 34 3. ssm mmh A hiker, who weighs 985 N, is strolling through the woods and crosses a small horizontal bridge. The bridge is uniform, weighs 360 N, and rests on two concrete supports, one at each end. He stops one-fifth of the way along the bridge. What is the magnitude of the force that a concrete support exerts on the bridge (a) at the near end and (b) at the far end? REASONING The drawing shows the bridge and the four forces that act on it: the upward force F exerted on the left end by the support, the force due to the weight W h of the hiker, the weight W b of the bridge, and the upward force F exerted on the right side by the support. Since the bridge is in equilibrium, the sum of the torques Σ τ = 0, and the sum of the forces in the about any axis of rotation must be zero ( ) vertical direction must be zero ( F y 0) determine the magnitudes of F and F. Σ =. These two conditions will allow us to F W h F +y +τ +x W b SOLUTION a. We will begin by taking the axis of rotation about the right end of the bridge. The torque produced by F is zero, since its lever arm is zero. When we set the sum of the torques equal to zero, the resulting equation will have only one unknown, F, in it. Setting the sum of the torques produced by the three forces equal to zero gives Σ τ = FL+ W 4 L + W L = 0 ( ) ( ) h 5 b Algebraically eliminating the length L of the bridge from this equation and solving for F gives ( ) ( ) 4 4 5 h b 5 F = W + W = 985 N + 360 N = 590 N b. Since the bridge is in equilibrium, the sum of the forces in the vertical direction must be zero: Σ F = F W W + F = Solving for F gives y h b 0 F = F+ Wh + Wb = 590 N + 985 N + 360 N = 00 N
Fall 07 PHYS 3 Chapter 9: 5, 0,, 3, 3, 34 3. Consult Multiple-Concept Example 0 to review an approach to problems such as this. A CD has a mass of 7 g and a radius of 6.0 cm. When inserted into a player, the CD starts from rest and accelerates to an angular velocity of rad/s in 0.80 s. Assuming the CD is a uniform solid disk, determine the net torque acting on it. REASONING The net torque Στ acting on the CD is given by Newton s second law for rotational motion (Equation 9.7) as Στ = Ι α, where I is the moment of inertia of the CD and α is its angular acceleration. The moment of inertia can be obtained directly from Table 9., and the angular acceleration can be found from its definition (Equation 8.4) as the change in the CD s angular velocity divided by the elapsed time. SOLUTION The net torque is Στ = Ι α. Assuming that the CD is a solid disk, its moment of inertia can be found from Table 9. as I = MR, where M and R are the mass and radius of the CD. Thus, the net torque is ( MR ) Σ τ = Iα = α The angular acceleration is given by Equation 8.4 as ( ) α = ω ω 0 / t, where ω and ω 0 are the final and initial angular velocities, respectively, and t is the elapsed time. Substituting this expression for α into Newton s second law yields ( MR ) α ( MR ) ω ω0 Σ τ = = t 3 rad/s 0 rad/s 4 = 7 0 kg 6.0 0 m 8.0 0 N m = 0.80 s 34. A ceiling fan is turned on and a net torque of.8 N m is applied to the blades. The blades have a total moment of inertia of 0. kg m. What is the angular acceleration of the blades? REASONING According to Newton s second law for rotational motion, Στ = Iα, the angular acceleration α of the blades is equal to the net torque Στ applied to the blades divided by their total moment of inertia I, both of which are known. SOLUTION The angular acceleration of the fan blades is τ.8 N m α = Σ = = 8. rad/s I 0. kg m (9.7)